What is the derivative of y=(8x^2+2x+6)/(sqrt(x)) ?

The derivative of y=(8x^2+2x+6)/(sqrt(x)) is (12x^2+x-3)/(xsqrt(x))

What is the first derivative of y=(8x^2+2x+6)/(sqrt(x)) ?

The first derivative of y=(8x^2+2x+6)/(sqrt(x)) is (12x^2+x-3)/(xsqrt(x))

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derivative of

y = \frac{8 x ^{2} + 2 x + 6}{x}

\frac{12 x ^{2} + x - 3}{x x}

Solution steps

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 6}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 8 x ^{2} + 2 x + 6 ) x - \frac{d}{d x} ( x ) ( 8 x ^{2} + 2 x + 6 )}{( x ) ^{2}}

\frac{d}{d x} (8 x^{2} + 2 x + 6) = 16 x + 2

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 6 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 6 )}{( x ) ^{2}} : \frac{12 x ^{2} + x - 3}{x x}

= \frac{12 x ^{2} + x - 3}{x x}

\frac{d}{d x} (1000 x e^{- 0.5 x})

\frac{\partial}{\partial x} (z^{3} ln (x y))

f (x) = - 2 x^{2} - 5 x + 2

\int_{0}^{2} 4^{- θ} d θ

\int (3 x^{2} - 1) d x

What is the derivative of y=(8x^2+2x+6)/(sqrt(x)) ?
The derivative of y=(8x^2+2x+6)/(sqrt(x)) is (12x^2+x-3)/(xsqrt(x))
What is the first derivative of y=(8x^2+2x+6)/(sqrt(x)) ?
The first derivative of y=(8x^2+2x+6)/(sqrt(x)) is (12x^2+x-3)/(xsqrt(x))